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Mathematics · Permutations and Combinations

JEE Main 2025 — 29 January, Morning Shift — Question 50

Let PP be the set of seven digit numbers with sum of their digits equal to 11 . If the numbers in P are formed by using the digits 1,2 and 3 only, then the number of elements in the set P is

  1. Option A:

    158

  2. Option B:

    173

  3. Option C:

    164

  4. Option D:

    161

    Correct

Answer: D

Step-by-step solution

We   need   7-tuples   (x1,…,x7), xi∈{1,2,3}, x1+⋯+x7=11.\text{We\; need\; 7-tuples\; }(x_1,\dots,x_7),\ x_i\in\{1,2,3\},\ x_1+\cdots+x_7=11. Put   yi=xi−1 (⇒yi∈{0,1,2}),y1+⋯+y7=11−7=4.\text{Put\; }y_i=x_i-1\ (\Rightarrow y_i\in\{0,1,2\}),\quad y_1+\cdots+y_7=11-7=4. Number   of   nonnegative   solutions   without   the   upper   bound   yi≤2 is   (4+7−17−1)=(106)=210.\text{Number\; of\; nonnegative\; solutions\; without\; the\; upper\; bound\; }y_i\le2 \text{ is\; } \binom{4+7-1}{7-1}=\binom{10}{6}=210. Count   solutions   with   a   given   yi≥3: set yi′=yi−3≥0, then   yi′+∑j≠iyj=1,\text{Count\; solutions\; with\; a\; given\; }y_i\ge3:\ \text{set }y_i'=y_i-3\ge0, \text{ then\; }y_i'+\sum_{j\ne i}y_j=1, which   has   (1+7−17−1)=(76)=7 solutions.\text{which\; has\; } \binom{1+7-1}{7-1}=\binom{7}{6}=7 \text{ solutions.} Since   two   yi≥3 is   impossible   (sum   is   4)  , inclusion–exclusion   gives :210−7⋅7=210−49=161.\text{Since\; two\; }y_i\ge3\text{ is\; impossible\; (sum\; is\; }4)\;, \text{ inclusion–exclusion\; gives } :210-7\cdot7=210-49=161. 161\boxed{161}

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Exam
JEE Main 2025
Paper
29 January, Morning Shift
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Problems based on both permutations and combinations

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